F(x) is continuous for all x.
Pick a point and show that f(x) is either negative or positive. Pick another point and show that f(x) is negative, if positive, or positive, if negative.
At x = 30, f(30) - 1000 = 900 + 10sin(30) - 1000 ≤ 0
Now, show at another point f(x) - 1000 is positive, and hence, there would be root between 30 and such point.
Let's pick 40.
At x = 40, f(40) - 1000 = 1600 + 10sin(40) - 1000 ≥ 0
Since f(x) - 1000 is continuous, there lies a root between 30 and 40, and hence, 30 ≤ c ≤ 40
<h3>
Answer: Choice D</h3>
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Explanation:
The inequality sign has an "or equal to", which means the boundary line will be solid. We can rule out choices B and C because they have dashed boundary lines.
A solid boundary line means that points on the boundary are part of the solution set.
Now let's see what happens when we plug in a point like (x,y) = (4,0). This will tell us how to shade the blue region.

This is false because -20 is not larger than -1. It's the other way around.
This tells us the point (4,0) is not in the blue shaded region, and it's not on the boundary line either. We can rule out choice A because of this.
The only thing left is choice D, which is the final answer. I recommend plugging a point from this region into the inequality to confirm we have a true statement.
4x + y = h
Rearrange the problem so your equation equals x, not h
y - h = -4x
Divide both sides by -4
-4y - 4h = x
Cheers!
A 0 will make the right decision
Answer:
TC (A) = 40x , TC (B) = 500 + 20x
Step-by-step explanation:
Let the number of students be = x
Hall A Total Cost
Relationship Equation, where TC (A) = f (students) = f (x) 40 per person (student) = 40x
Hall B Total Cost
Relationship Equation, where TC (B) = f (students) = f (x) 500 fix fee & 20 per person (student) = 500 + 20x